Ngô Quốc Anh

July 31, 2009

A property of the essentially bounded function


Let E be a subset of \mathbb R^n with |E| < \infty in Lebesgue sense. Suppose f \in L^\infty(E) and \| f\|_\infty > 0. Set

\displaystyle {a_n} = \int_E {{{\left| f \right|}^n}}

for n=1,2,3,... Show that

\displaystyle\mathop {\lim }\limits_{n \to \infty } \frac{{{a_{n + 1}}}} {{{a_n}}} = {\left\| f \right\|_\infty }.

Solution. For any \alpha with 0<\alpha < \|f\|_\infty, let

\displaystyle{E_\alpha } = \left\{ {x \in E: f\left( x \right) \geqslant \alpha } \right\}

and

\displaystyle {F_\alpha } = E\backslash {E_\alpha }

then |E_\alpha|>0. For any k \in \mathbb N, by the Dominated Convergence Theorem,

\displaystyle\mathop{\lim }\limits_{n\to\infty }\left({\dfrac{{\int_{{F_\alpha }}{{{\left| f\right|}^{n+k}}}}}{{\int_{{E_\alpha }}{{{\left| f\right|}^{n}}}}}}\right)\leqslant\underbrace{\mathop{\lim }\limits_{n\to\infty }\frac{1}{{\left|{{E_\alpha }}\right|}}\int_{{F_\alpha }}{{{\left|{\frac{f}{\alpha }}\right|}^{n}}\left\| f\right\|_\infty^{k}}}_{0}.

Hence

\displaystyle\mathop{\lim\inf }\limits_{n\to\infty }\left({\frac{{\int_{E}{{{\left| f\right|}^{n+1}}}}}{{\int_{E}{{{\left| f\right|}^{n}}}}}}\right)\geqslant\mathop{\lim\inf }\limits_{n\to\infty }\left({\frac{{\alpha\int_{{E_\alpha }}{{{\left| f\right|}^{n}}}+\int_{{F_\alpha }}{{{\left| f\right|}^{n+1}}}}}{{\int_{{E_\alpha }}{{{\left| f\right|}^{n}}}+\int_{{F_\alpha }}{{{\left| f\right|}^{n}}}}}}\right) =\alpha .

Letting \alpha \nearrow {\left\| f \right\|_\infty }, we get that

\displaystyle\mathop{\lim }\limits_{n\to\infty }\left({\frac{{\int_{E}{{{\left| f\right|}^{n+1}}}}}{{\int_{E}{{{\left| f\right|}^{n}}}}}}\right) ={\left\| f\right\|_\infty }.

As an application, if we put a_0 = 1, then from

\displaystyle {a_{n + 1}} = \frac{{{a_1}}} {{{a_0}}}.\frac{{{a_2}}} {{{a_1}}} \cdots\frac{{{a_{n + 1}}}} {{{a_n}}}

we deduce that

\displaystyle\mathop{\lim }\limits_{n\to\infty }\sqrt[n]{{{a_{n}}}}=\mathop{\lim }\limits_{n\to\infty }\frac{{{a_{n+1}}}}{{{a_{n}}}}={\left\| f\right\|_\infty }.

In other words,

\displaystyle\mathop{\lim }\limits_{n\to\infty }{\left({\int_{E}{{{\left| f\right|}^{n}}}}\right)^{\frac{1}{n}}}={\left\| f\right\|_\infty }.

Picard’s Theorem + Hadamard’s Theorem = ?


Question. Let f be an entire non-constant function that satisfies the functional equation

\displaystyle f(1 - z) = 1 - f(z)

for all z \in \mathbb C. Show that f(\mathbb C) = \mathbb C.

Solution. Assume by contradiction, then W.L.O.G. by the Picard’s Theorem we can assume that f misses a \in \mathbb C. By the Hadamard’s Theorem,

\displaystyle f(z)-a = e^{p(z)}

for some polynomial p. Therefore,

\displaystyle f(z) = a +e^{p(z)}

for all z \in \mathbb C. From the fact that

\displaystyle f(1-z)=1-f(z)

we get

\displaystyle \underbrace{a+{e^{p\left({1-z}\right)}}}_{f\left({1-z}\right)}=\underbrace{1-\left({a+{e^{p\left( z\right)}}}\right)}_{1-f\left( z\right)}

which yields

\displaystyle {e^{p\left( z \right)}} = 2a - 1 + {e^{p\left( {1 - z} \right)}}.

Put z=0 and z=1, we obtain

\displaystyle {e^{p\left( 0 \right)}} = 2a - 1 + {e^{p\left( 1 \right)}}, \quad {e^{p\left( 1 \right)}} = 2a - 1 + {e^{p\left( 0 \right)}}.

Hence

\displaystyle {e^{p\left( 0 \right)}} = 2\left( {2a - 1} \right) + {e^{p\left( 0 \right)}}

which implies a=\frac{1}{2}. From the identity f(1-z)=1-f(z) put z=\frac{1}{2} we then deduce that

\displaystyle f\left( {\frac{1} {2}} \right) = \underbrace {\frac{1} {2}}_a

a contradiction.

Note: I think I should post some applications of the Hadamard’s Theorem.

Corrigendum to the previous proof. Assume by contradiction, then f takes all values except possibly some a. Also notethat f(1/2) = 1/2. So we can assume a \neq 1/2 which means that 1-a \ne a. It then does take the value 1-a, and hence takes the value a. The proof is now complete.

I thank Xu Wei Biao for pointing out a mistake in the previous solution.

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